The equation of the circle with centre at (4, 3) and touching the line 5x–12y–10 = 0 is 
 

$$\begin{gathered} \mathrm{centre}\left(\mathrm{C}\right)=(4, 3) \\ Given line 5x-12y-10=0 \\ \sin ce it is a \tan gent then \\ \mathrm{r}=\mathrm{d} \\ \Rightarrow \mathrm{r}=\frac{\left|20-36-10\right|}{\sqrt{169}} \\ \Rightarrow \mathrm{r}=\frac{26}{13}=2 \\ Now circle is {(\mathrm{x}-4)}^2+{(\mathrm{y}-3)}^2=4 \end{gathered}$$